Question: Solve for $q$, $ \dfrac{3q - 3}{q + 3} = \dfrac{1}{9} $
Solution: Multiply both sides of the equation by $q + 3$ $ 3q - 3 = \dfrac{q + 3}{9} $ Multiply both sides of the equation by $9$ $ 9(3q - 3) = q + 3 $ $27q - 27 = q + 3$ $26q - 27 = 3$ $26q = 30$ $q = \dfrac{30}{26}$ Simplify. $q = \dfrac{15}{13}$